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Past Questions — 2010
A complete binary tree with the property that the value at each node is at least as large as the values at its children in known as
A)
binary search tree
B)
AVL-tree
C)
completely balanced tree
D)
heap
The question describes the fundamental properties of a Max-Heap.
1. **Structural Property**: The question specifies a 'complete binary tree'. A complete binary tree is one where every level, except possibly the last, is completely filled, and all nodes are as far left as possible. This is a requirement for a binary heap.
2. **Heap Property**: The condition that 'the value at each node is at least as large as the values at its children' is the definition of the Max-Heap property. In a Max-Heap, for any given node I, the value of I is greater than or equal to the values of its children.
Why other options are incorrect:
- **a. binary search tree**: In a BST, the left child must be smaller than the parent, and the right child must be larger. It does not require the tree to be complete, nor does it require the parent to be larger than both children.
- **b. AVL-tree**: An AVL tree is a self-balancing binary search tree. While it maintains a height balance, it follows BST ordering rules, not the heap property.
- **c. completely balanced tree**: This refers to the structural balance of a tree (where the heights of subtrees differ by at most 1), but it does not define a specific relationship between the values of parents and children.
What is the time required to search an element in a linked list of length n?
A)
O (log2 n)
B)
O (n)
C)
O (1)
D)
O (n2 )
In a linked list, elements are not stored in contiguous memory locations; instead, each node contains a pointer to the next node. Because there is no direct index-based access (unlike an array), the only way to search for a specific element is to perform a 'Linear Search'. This involves starting at the head node and traversing the list sequentially one node at a time until the target element is found or the end of the list is reached. In the worst-case scenario (where the element is at the very end of the list or not present at all), the algorithm must visit all 'n' nodes. Therefore, the time complexity is proportional to the number of elements, which is expressed in Big O notation as O(n).
Why other options are incorrect:
- O(log2 n): This is the complexity of a Binary Search. Binary search requires random access to the middle element, which is not possible in a standard linked list.
- O(1): This is constant time. This would only be possible if we had a direct pointer to the element we were searching for.
- O(n2): This is quadratic time, typically associated with nested loops (like Bubble Sort). Searching a list only requires a single pass.
A graph G with n nodes is bipartite if it contains
A)
n edges
B)
a cycle of odd length
C)
no cycle of odd length
D)
n2 edges
A bipartite graph is a graph whose vertices can be divided into two disjoint and independent sets $U$ and $V$ such that every edge connects a vertex in $U$ to one in $V$.
1. **Fundamental Theorem**: A graph is bipartite if and only if it contains no odd cycles. This is a well-known theorem in graph theory (Kőnig's Theorem).
2. **Reasoning**: In a bipartite graph, any path that starts in set $U$ and returns to set $U$ must traverse an even number of edges (U -> V -> U). Therefore, any cycle (a path that starts and ends at the same vertex) must have an even length. If a graph contains a cycle of odd length (e.g., a triangle), it is impossible to partition the vertices into two sets without having an edge connect two vertices within the same set.
**Analysis of Options:**
- **Option a (n edges):** The number of edges does not determine if a graph is bipartite. A bipartite graph can have anywhere from 0 to $n^2/4$ edges.
- **Option b (a cycle of odd length):** This is the exact opposite of the requirement. If a graph has an odd cycle, it is NOT bipartite.
- **Option c (no cycle of odd length):** This is the correct mathematical definition/property of a bipartite graph.
- **Option d (n^2 edges):** A simple graph with $n$ nodes can have at most $n(n-1)/2$ edges. $n^2$ edges are impossible for a simple graph, and the number of edges alone does not define bipartiteness.
An instruction with an indexed operand having address field with all 0’s bit is effectively a
A)
register indirect mode of operation
B)
register direct mode of operation
C)
memory indirect mode of operation
D)
base-register addressing mode of operation
In indexed addressing mode, the effective address (EA) of the operand is calculated by adding the contents of an index register (X) to the address field (displacement/offset) provided in the instruction. The formula is: EA = Content of Index Register + Address Field. If the address field consists of all 0's bits, the formula becomes: EA = Content of Index Register + 0, which simplifies to EA = Content of Index Register. This means the instruction is using the value stored in the register as the memory address to access the operand. This is the definition of Register Indirect Addressing, where the register holds the pointer to the memory location of the operand.
Why other options are incorrect:
- b. Register Direct: In this mode, the operand itself is stored in the register, not at a memory address pointed to by the register.
- c. Memory Indirect: This involves accessing memory twice (once for the address and once for the operand), which is not the case here.
- d. Base-register addressing: While similar to indexed addressing, base-register addressing typically uses a base register to point to the start of a segment and a non-zero offset to find a specific element; if the offset is zero, it still functions as a register indirect access, but the specific mechanism described (indexed operand with 0 offset) is the classic definition of transforming indexed mode into register indirect mode.
What is the time complexity of linear search algorithm over an array of n elements?
A)
O (log2 n)
B)
O (n)
C)
O (n log2 n)
D)
O (n2 )
Linear search is the simplest searching algorithm. It works by starting at the first element of an array and comparing the target value with each element sequentially until a match is found or the end of the array is reached.
1. **Best Case Complexity**: O(1) - This occurs if the target element is the first element of the array.
2. **Worst Case Complexity**: O(n) - This occurs if the target element is the last element of the array or is not present at all. In this case, the algorithm must perform 'n' comparisons for an array of size 'n'.
3. **Average Case Complexity**: O(n) - On average, the element will be found in the middle (n/2), which simplifies to O(n) in Big O notation.
Analysis of other options:
- O(log2 n) is the complexity of Binary Search, which requires the array to be sorted.
- O(n log2 n) is typical for efficient sorting algorithms like Merge Sort or Quick Sort.
- O(n^2) is typical for simple sorting algorithms like Bubble Sort or Selection Sort.
The goal of hashing is to produce a search that takes
A)
O(1) time
B)
O(n2) time
C)
O(log n) time
D)
O(n log n) time
Hashing is a technique used to uniquely identify a specific object from a group of similar objects using a hash function. The primary goal of hashing is to map a key to a specific index in a hash table such that the time required to insert, delete, or search for an element is independent of the number of elements (n) stored in the table.
1. **O(1) - Constant Time**: In an ideal scenario (with a good hash function and minimal collisions), the hash function computes the address of the data instantly, allowing for search operations to be performed in constant time, denoted as O(1). This is the fundamental objective of using a hash table over other data structures.
2. **O(log n) - Logarithmic Time**: This is the typical search time for balanced binary search trees (BSTs) or binary search on sorted arrays. Hashing aims to be faster than this.
3. **O(n log n)**: This is common for efficient sorting algorithms (like Merge Sort or Heap Sort), not for searching in a hash table.
4. **O(n²)**: This represents quadratic time, which is highly inefficient and typical of nested loops in simple algorithms (like Bubble Sort), not a search goal.
Which data structure is needed to convert infix notations to postfix notations?
A)
Linear list
B)
Queue
C)
Tree
D)
Stack
To convert an infix expression (e.g., A + B) to a postfix expression (e.g., AB+), the Shunting-yard algorithm is typically used. This algorithm requires a Stack to temporarily hold operators and parentheses. The stack is used to manage the precedence and associativity of operators: when an operator is encountered, it is pushed onto the stack if it has higher precedence than the operator currently on top; otherwise, the top operator is popped and added to the output string.
Why other options are incorrect:
- a. Linear list: While a stack is a type of linear list, 'Linear list' is too generic. A simple list does not inherently provide the Last-In-First-Out (LIFO) mechanism required for operator precedence management.
- b. Queue: A queue follows First-In-First-Out (FIFO), which would process operators in the order they appear, ignoring mathematical precedence rules.
- c. Tree: While an expression can be represented as an Expression Tree, the process of converting infix to postfix is fundamentally handled by a stack. A tree is a result or a different representation, not the primary data structure used for the conversion algorithm itself.
A sort which uses the binary tree concept such that any number is larger than all the numbers in the subtree below it is called
A)
selection sort
B)
insertion sort
C)
heap sort
D)
quick sort
The question describes the fundamental property of a Max-Heap. A Heap is a specialized complete binary tree. In a Max-Heap, for any given node, the value of that node is greater than or equal to the values of its children (the subtree below it). Heap Sort utilizes this data structure by building a max-heap and repeatedly extracting the maximum element (the root) to sort the array.
Why other options are incorrect:
- a. Selection Sort: This is a comparison-based sort that repeatedly finds the minimum element from the unsorted part and puts it at the beginning. It does not use a tree structure.
- b. Insertion Sort: This builds the final sorted array one item at a time by shifting elements. It does not use a tree structure.
- d. Quick Sort: While Quick Sort uses a divide-and-conquer approach that can be visualized as a recursion tree, it does not maintain a binary tree where a parent is always larger than its subtree; it partitions elements based on a pivot.
A carbabion has
A)
SP3-hybrid carbon
B)
SP2-hybrid carbon
C)
SP-hybrid carbon
D)
dSP2-hybrid carbon
A carbanion is a chemical species in which a carbon atom possesses a negative charge and a lone pair of electrons. To determine the hybridization of the central carbon in a simple carbanion (like the methyl anion, CH3-), we look at the steric number: Steric Number = (Number of bonded atoms) + (Number of lone pairs). For a typical carbanion, the carbon is bonded to three other atoms (or hydrogens) and possesses one lone pair. Steric Number = 3 + 1 = 4. A steric number of 4 corresponds to sp3 hybridization. The geometry is typically pyramidal, similar to ammonia (NH3). While resonance or adjacent pi-systems can change the hybridization to sp2 in specific cases (like allyl or benzyl carbanions), the general definition of a carbanion refers to the sp3 state. Option b (sp2) and Option c (sp) are incorrect for a standard carbanion, and Option d (dsp2) is incorrect as carbon is a second-period element and does not have d-orbitals available for hybridization.
Nitrene can be detected by which one of the following spectroscopic method?
A)
ESR
B)
IR
C)
NMR
D)
UV-VIS
Nitrenes are the nitrogen analogues of carbenes. They are highly reactive intermediates characterized by having a divalent nitrogen atom with six valence electrons. Most nitrenes exist in a triplet ground state, meaning they possess two unpaired electrons. Electron Spin Resonance (ESR), also known as Electron Paramagnetic Resonance (EPR), is the specialized spectroscopic technique used to detect species with unpaired electrons (paramagnetic species). Because triplet nitrenes are paramagnetic, ESR is the primary and most effective method for their detection and characterization.
Why other options are incorrect:
- b. IR (Infrared Spectroscopy): Used for identifying functional groups based on bond vibrations; while it can detect some species, it is not the primary tool for identifying the electronic state of short-lived paramagnetic intermediates like nitrenes.
- c. NMR (Nuclear Magnetic Resonance): Used for diamagnetic species. Paramagnetic species like triplet nitrenes cause extreme line broadening and shifting in NMR, making it unsuitable for their direct detection.
- d. UV-VIS (Ultraviolet-Visible Spectroscopy): Used for electronic transitions in molecules. While nitrenes may absorb light, UV-VIS cannot specifically confirm the presence of the unpaired electrons that define the nitrene's reactive nature as effectively as ESR.
Condensation of two molecules of ethylacetate in the presence of sodium ethoxide gives ethylacetoacetic ester. This reaction is known as
A)
Claisen ester condensation
B)
Aldol condensation
C)
Dieckmann condensation
D)
None of these
The reaction described is the condensation of two molecules of an ester (ethyl acetate) in the presence of a strong base (sodium ethoxide) to form a $\beta$-keto ester (ethyl acetoacetate). This specific reaction is known as the Claisen condensation.
Step-by-step mechanism:
1. The ethoxide ion ($ ext{CH}_3 ext{CH}_2 ext{O}^-$) abstracts an $\alpha$-hydrogen from one molecule of ethyl acetate to form an enolate ion.
2. This enolate ion acts as a nucleophile and attacks the carbonyl carbon of a second molecule of ethyl acetate.
3. An ethoxide ion is expelled as a leaving group, resulting in the formation of ethyl acetoacetate.
Analysis of other options:
- b. Aldol condensation: This involves the condensation of aldehydes or ketones to form $\beta$-hydroxy aldehydes or ketones, not esters.
- c. Dieckmann condensation: This is an intramolecular version of the Claisen condensation, where a dicarboxylic ester cyclizes to form a cyclic $\beta$-keto ester.
- d. None of these: Incorrect, as option 'a' is the precise name for this reaction.
The chlorination of methane in the presence of ultraviolet light is an example of
A)
Electrophilic substitution
B)
Nucleophilic substitution
C)
free-radical substitution
D)
Nucleophilic addition
The chlorination of methane (CH₄ + Cl₂ → CH₃Cl + HCl) occurs via a free-radical chain mechanism, which is triggered by ultraviolet (UV) light. The process consists of three main stages:
1. **Initiation**: UV light provides the energy to cause the homolytic fission of the Cl-Cl bond, creating two chlorine free radicals (Cl•).
2. **Propagation**: A chlorine radical attacks a methane molecule to form a methyl radical (•CH₃) and HCl. The methyl radical then reacts with another Cl₂ molecule to form chloromethane (CH₃Cl) and regenerate a chlorine radical.
3. **Termination**: Two radicals collide and combine to form a stable molecule, ending the chain.
Why other options are incorrect:
- **Electrophilic substitution (a)**: This typically occurs in aromatic systems (like benzene) where an electrophile replaces a hydrogen atom.
- **Nucleophilic substitution (b)**: This involves a nucleophile attacking an electron-deficient carbon (common in alkyl halides), not the initial chlorination of an alkane.
- **Nucleophilic addition (d)**: This occurs in compounds with multiple bonds, such as carbonyl groups (C=O), where a nucleophile adds to the unsaturated center.
Which one of the following primary alkyl bromide give nucleophilic substitution by SNI mechanism?
A)
(CH3)3 C CH2Br
B)
CH3CH2Br
C)
CH3Br
D)
CH3CH2CH2Br
The question asks which primary alkyl bromide can undergo nucleophilic substitution via the SN1 mechanism.
1. **SN1 Mechanism Basics**: The SN1 mechanism involves the formation of a carbocation intermediate. The rate-determining step is the ionization of the C-X bond. Therefore, the stability of the resulting carbocation determines whether the SN1 pathway is feasible.
2. **Primary Carbocations**: Generally, primary alkyl halides (like options b, c, and d) do not undergo SN1 reactions because primary carbocations are highly unstable. They typically react via the SN2 mechanism.
3. **Analyzing Option (a) - Neopentyl Bromide**: (CH3)3C-CH2Br is a primary alkyl bromide. While the initial carbocation formed would be primary (and unstable), it is immediately adjacent to a quaternary carbon. This allows for a simultaneous or rapid '1,2-methyl shift' (Wagner-Meerwein rearrangement) to convert the unstable primary carbocation into a highly stable tertiary carbocation ((CH3)2C(+)-CH2CH3).
4. **Comparison**:
- Option (b) Ethyl bromide: Forms a primary carbocation (unstable).
- Option (c) Methyl bromide: Forms a methyl cation (extremely unstable).
- Option (d) n-Propyl bromide: Forms a primary carbocation (unstable).
- Option (a) Neopentyl bromide: Can undergo SN1 because the resulting primary carbocation rearranges instantly to a stable tertiary carbocation, making the overall process energetically favorable compared to the other primary options.
OsO4 is a valuable reagent in organic synthesis. It is a
A)
reducing agent
B)
oxidising agent
C)
dehydrating agent
D)
hydroxylating agent
OsO4 (Osmium tetroxide) is a powerful oxidizing agent used primarily in organic chemistry for the syn-hydroxylation of alkenes. In this reaction, OsO4 reacts with a carbon-carbon double bond (C=C) to form a cyclic osmate ester, which is then hydrolyzed to produce a cis-1,2-diol (vicinal diol). Since the oxidation state of the carbon atoms increases (from +1 or 0 in the alkene to +1 in the alcohol) and the osmium atom is reduced from +8 to a lower oxidation state during the process, OsO4 acts as an oxidizing agent. While it does 'hydroxylate' the molecule (adding hydroxyl groups), 'oxidising agent' is the standard chemical classification for its role in this redox reaction. Option (a) is incorrect because it gains electrons/reduces the substrate; option (c) is incorrect as it does not remove water; option (d) is a descriptive term for the result, but 'oxidising agent' is the correct chemical functional classification.
A triplet carbene has two electrons in
A)
two different orbitals
B)
same orbital
C)
two different orbits
D)
None of these
Carbenes are neutral species containing a divalent carbon atom with six valence electrons. They exist in two electronic states: singlet and triplet.
1. **Singlet Carbene**: In a singlet carbene, the two non-bonding electrons are paired (opposite spins) and occupy the same sp2 hybrid orbital.
2. **Triplet Carbene**: In a triplet carbene, the two non-bonding electrons have parallel spins (same spin) and occupy two different orbitals (typically one sp2 hybrid orbital and one p-orbital). According to Hund's Rule, electrons occupy degenerate or near-degenerate orbitals singly with parallel spins to minimize electron-electron repulsion.
Therefore, a triplet carbene has its two electrons in two different orbitals. Option (b) describes a singlet carbene, and option (c) uses the term 'orbits', which is an outdated Bohr-model term and incorrect in the context of quantum mechanical orbitals.
The reaction of H2C=CHBr with sodamide gives
A)
Acetylene
B)
Ethane
C)
Ethene
D)
None of these
The reaction described is a dehydrohalogenation reaction. The starting material is vinyl bromide (H2C=CHBr). Sodamide (NaNH2) is an extremely strong base. When vinyl bromide reacts with sodamide, the strong base abstracts a proton from the carbon atom adjacent to the carbon bearing the bromine (the beta-carbon relative to the halogen in the context of the resulting triple bond). This leads to the elimination of HBr (as NaBr and NH3), resulting in the formation of a carbon-carbon triple bond. The chemical equation is: H2C=CHBr + NaNH2 → HC≡CH (Acetylene) + NaBr + NH3.
Analysis of options:
- a. Acetylene: Correct. The elimination of HBr from vinyl bromide yields ethyne (acetylene).
- b. Ethane: Incorrect. Ethane is a saturated hydrocarbon (alkane) and would require hydrogenation, not elimination.
- c. Ethene: Incorrect. Ethene is the starting alkene structure (without the bromine); the reaction removes HBr to increase the degree of unsaturation.
- d. None of these: Incorrect, as option 'a' is the correct product.
Reformatsky reaction consists in reacting aldehydes and ketones with a-bromoesters in the presence of
A)
Magnesium
B)
Zinc
C)
Cadmium
D)
Zinc amalgam
The Reformatsky reaction is a classic organic chemical reaction used to synthesize $\beta$-hydroxy esters. It involves the condensation of an aldehyde or a ketone with an $\alpha$-haloester (typically an $\alpha$-bromoester) in the presence of metallic zinc.
Step-by-step mechanism:
1. **Formation of Organozinc Reagent**: The zinc metal inserts itself into the carbon-bromine bond of the $\alpha$-bromoester to form an organozinc bromide (an enolate). This species is less reactive than a Grignard reagent, which prevents it from attacking the ester group of another molecule.
2. **Nucleophilic Attack**: The organozinc reagent then attacks the carbonyl carbon of the aldehyde or ketone.
3. **Hydrolysis**: Subsequent acid hydrolysis yields the final $\beta$-hydroxy ester.
Comparison with options:
- **Magnesium (a)**: Used in Grignard reactions. If magnesium were used with $\alpha$-bromoesters, the resulting reagent would be too reactive and would attack the ester group of the starting material (self-condensation).
- **Zinc (b)**: The correct metal required for the specific reactivity of the Reformatsky reaction.
- **Cadmium (c)**: Not the standard metal for this reaction.
- **Zinc amalgam (d)**: While zinc amalgam is used in other reactions (like the Clemmensen reduction), the standard Reformatsky reaction specifically utilizes metallic zinc.
The reaction of an alkene with NBS gives product A. A is
A)
Allylic bromide
B)
Vinylic bromide
C)
Gem-dibromide
D)
None of these
NBS (N-Bromosuccinimide) is a reagent specifically used for the radical substitution of bromine at the allylic position of an alkene. The mechanism involves the generation of a low concentration of bromine radicals (Br•). These radicals abstract a hydrogen atom from the allylic position (the carbon atom adjacent to the double bond) to form a resonance-stabilized allylic radical. This radical then reacts with Br2 (generated in situ) to yield an allylic bromide.
Why other options are incorrect:
- b. Vinylic bromide: Bromination of the vinylic position (the carbons of the double bond) typically requires different conditions or reagents and is energetically less favorable than allylic substitution.
- c. Gem-dibromide: This refers to two bromine atoms on the same carbon. NBS is used for mono-bromination at the allylic position, not for adding two bromine atoms to a single carbon.
- d. None of these: Incorrect, as option 'a' is the standard result of this reaction.
Which one of the following compound will give Cannizzaro’s reaction?
A)
CH3CHO
B)
C6H5 CH2CHO
C)
(CH3)3 C-CHO
D)
CH3CH2CHO
Cannizzaro's reaction is a chemical reaction where an aldehyde that lacks an alpha-hydrogen (α-hydrogen) undergoes self-oxidation and reduction (disproportionation) in the presence of a strong base to form an alcohol and a salt of a carboxylic acid.
Let's analyze the options:
1. CH3CHO (Acetaldehyde): Has three alpha-hydrogens on the methyl group. It will undergo Aldol condensation instead of Cannizzaro's reaction.
2. C6H5CH2CHO (Phenylacetaldehyde): Has two alpha-hydrogens on the carbon adjacent to the carbonyl group. It will undergo Aldol condensation.
3. (CH3)3C-CHO (Pivalaldehyde/Trimethylacetaldehyde): The alpha-carbon is a quaternary carbon bonded to three methyl groups; it has NO alpha-hydrogens. Therefore, it is the only compound among the options that can undergo Cannizzaro's reaction.
4. CH3CH2CHO (Propionaldehyde): Has two alpha-hydrogens on the second carbon. It will undergo Aldol condensation.
Since only (CH3)3C-CHO lacks alpha-hydrogens, it is the correct answer.
A nucleoplilic reagent will readily attack
A)
Ethylene
B)
Ethanal
C)
Ethanol
D)
Ethyl amine
A nucleophilic reagent is an electron-rich species (nucleophile) that seeks an electron-deficient center (electrophile) to donate a pair of electrons and form a chemical bond.
1. **Ethanal (CH₃CHO)**: This is an aldehyde. The carbonyl carbon (C=O) is highly electrophilic because oxygen is more electronegative than carbon, creating a partial positive charge (δ+) on the carbon atom. Additionally, the planar geometry of the carbonyl group allows easy access for a nucleophile to attack. Therefore, nucleophiles readily attack the carbonyl carbon of ethanal.
2. **Ethylene (CH₂=CH₂)**: This is an alkene. The double bond is a region of high electron density (rich in π-electrons). Since nucleophiles are also electron-rich, they are repelled by the double bond. Alkenes typically undergo electrophilic addition, not nucleophilic attack.
3. **Ethanol (CH₃CH₂OH)**: In ethanol, the carbon atoms are not sufficiently electrophilic. While the carbon attached to oxygen has a slight positive character, it is not as reactive as a carbonyl carbon, and the hydroxyl group is not a good leaving group under standard conditions.
4. **Ethyl amine (CH₃CH₂NH₂)**: The nitrogen atom in amines has a lone pair of electrons, making the molecule itself a nucleophile. A nucleophilic reagent will not attack another nucleophilic center due to electrostatic repulsion.
Conclusion: Ethanal is the only option containing a strong electrophilic center (the carbonyl carbon), making it the correct target for a nucleophilic reagent.
Zeigler-Natta catalyst is
A)
K[Pt Cl3(C2H4)]
B)
(Pb3P)3 RhCl
C)
Al2(C2H5)6 + TiCl4
D)
Fe(C5H5)2
The Ziegler-Natta catalyst is a coordination catalyst used primarily for the polymerization of 1-alkenes (such as ethylene and propylene) to produce linear, high-density polymers. It typically consists of two components: a transition metal compound (the catalyst) and an organometallic compound (the co-catalyst).
1. Analysis of Option (c): The combination of Titanium tetrachloride (TiCl4) and Triethylaluminium [Al(C2H5)3, which exists as a dimer Al2(C2H5)6] is the classic composition of the Ziegler-Natta catalyst. The TiCl4 acts as the transition metal source, and the Al-alkyl acts as the activating agent.
2. Analysis of other options:
- Option (a): K[PtCl3(C2H4)] is Zeise's salt, the first organometallic compound discovered, but it is not a Ziegler-Natta catalyst.
- Option (b): This represents a Rhodium-based complex, likely related to catalysts used in hydroformylation (like Wilkinson's catalyst), not Ziegler-Natta polymerization.
- Option (d): Fe(C5H5)2 is Ferrocene, a prototypical metallocene. While some modern 'metallocene catalysts' are used in polymerization, the standard 'Ziegler-Natta catalyst' specifically refers to the Ti/Al system.
The spectra which corresponds to the wave number 50,000 cm-1 is
A)
NMR
B)
Visible
C)
Microwave
D)
UV
To determine which region of the electromagnetic spectrum corresponds to a wave number of 50,000 cm⁻¹, we use the relationship between wave number (ν̄) and wavelength (λ):
1. Formula: ν̄ = 1 / λ
2. Calculation: λ = 1 / ν̄ = 1 / 50,000 cm⁻¹ = 0.00002 cm
3. Conversion to nanometers (nm): 0.00002 cm × 10⁷ nm/cm = 200 nm
Now, we compare this wavelength to the standard regions of the electromagnetic spectrum:
- Microwave: Wavelengths typically range from 1 mm to 1 m (very low wave numbers).
- NMR (Radio waves): Wavelengths typically range from 1 m to 100 km (extremely low wave numbers).
- Visible: Wavelengths range from approximately 400 nm to 700 nm (wave numbers approx. 14,000 to 25,000 cm⁻¹).
- UV (Ultraviolet): Wavelengths range from approximately 10 nm to 400 nm (wave numbers approx. 25,000 to 100,000 cm⁻¹).
Since 200 nm (or 50,000 cm⁻¹) falls squarely within the Ultraviolet (UV) range, option (d) is the correct answer.
Friedel-Crafts reaction of bromo benzene with methyl iodide gives
A)
O–Bromotoluene
B)
p–Bromotoluene
C)
O– and p–Bromotoluene
D)
m–Bromotoluene
1. The reaction described is a Friedel-Crafts Alkylation, where bromobenzene reacts with methyl iodide (CH3I) in the presence of a Lewis acid catalyst (like AlCl3) to introduce a methyl group onto the benzene ring.
2. In bromobenzene, the bromine atom is already attached to the ring. Bromine is a halogen, and in electrophilic aromatic substitution reactions, halogens act as ortho/para directing groups.
3. This is because the lone pairs of electrons on the bromine atom can stabilize the intermediate sigma complex through resonance, specifically when the electrophile (the methyl cation, CH3+) attacks the ortho or para positions.
4. Therefore, the methyl group will be directed to the positions ortho (2nd position) and para (4th position) relative to the bromine atom.
5. The resulting products are o-bromotoluene and p-bromotoluene. While the para isomer is often the major product due to steric hindrance at the ortho position, both are formed in the reaction.
6. Option (a) and (b) are partially correct but incomplete, and option (d) is incorrect because bromine is not a meta-directing group.
Which proton of Br2CH-CH2Br give triplet splitting in NMR
A)
C - H
B)
- CH2
C)
Both of these
D)
None of these
CHBr2-CH2Br: CH coupled to CH2 (2H) => n+1=3 triplet; CH2 coupled to 1H => doublet, so triplet is CH proton.
Formaldehyde reacts with NaOH (50%). Which one pair of product will be obtained?
A)
CH3COOH + CH3OH
B)
CH3OH + CH3COONa
C)
CH3COOH + C2H5OH
D)
HCOONa + CH3OH
The reaction described is the Cannizzaro reaction. Formaldehyde (HCHO) is an aldehyde that lacks alpha-hydrogen atoms. When such aldehydes react with a concentrated strong base like NaOH (50%), they undergo a self oxidation-reduction (disproportionation) reaction. In this process, one molecule of formaldehyde is reduced to an alcohol (methanol, CH3OH), and another molecule is oxidized to the salt of a carboxylic acid (sodium formate, HCOONa). The balanced chemical equation is: 2HCHO + NaOH → HCOONa + CH3OH. Option (a) and (c) are incorrect because the reaction occurs in a basic medium, so the product is a salt (HCOONa), not a free acid (HCOOH). Option (b) is incorrect because it suggests the formation of acetate (CH3COONa), which would require an acetaldehyde starting material, not formaldehyde.