JEE Advanced · Chemistry · 24. Haloalkanes & Haloarenes

Match the following
Column I Column II
(A) $CH _3- CHBr - CD _3$ on treatment with alc. KOH gives $CH _2= CH - CD _3$ as a major product. (P) E1 reaction

(B) $Ph - CHBr - CH _3$ reacts faster than $Ph - CHBr - CD _3$ (Q) E2 reaction
(C) $Ph - CH _2- CH _2 Br$ on treatment with $C _2 H _5 OD / C _2 H _5 O ^{-}$gives $Ph - CD = CH _2$ as the major product (R) El cb reaction
(D) $PhCH _2 CH _2 Br$ and $PhCH _2 CH _2 Br$ react with same rate (S) First order reaction

A (A) Q, (B) P, (C) R,S, (D) Q,S
B (A) S, (B) R, (C) R, (D) P,S
C (A) R, (B) Q, (C) S, (D) Q,S
D (A) Q, (B) Q, (C) R,S, (D) P,S

Verified Solution

Answer & Solution

Correct Answer

(D) (A) Q, (B) Q, (C) R,S, (D) P,S

Step-by-step Solution

Detailed explanation

(A) Match with (Q)

The formation of $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CD}_3$ can be explained on the basis of the fact that $\mathrm{C}-\mathrm{D}$ bond is much stronger than $\mathrm{C}-\mathrm{H}$ bond.
(B) match with (Q)
Reactivity of $\mathrm{PhCHBrCH}_3$ is greater than
$\mathrm{Ph} \mathrm{CHBrCD}_3$ because $\mathrm{C}-\mathrm{D}$ bond is more stronger than $\mathrm{C}-\mathrm{H}$ bond.
(C) match with (R) and (S)

In the step (II), a slow unimolecular elimination occurs in the conjugate base of the reactant and hence this mechanism is called $E_1 C B$ or carbanion mechanism. Since step (I) must be reversible, if ethanol containing $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD}$ is used as solvent, it would be expected that the original bromide would incorporate deuterium (D).
$\text { (D) Match with (P) and (S) }$
Step I. $\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br} \stackrel{\text { Slow }}{\longrightarrow} \mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}} \mathrm{H}_2+\mathrm{Br}^{-}$
Step II. $\mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}_2} \mathrm{H}_2 \stackrel{\text { Fast }}{\longrightarrow} \mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_2+\mathrm{H}^{-}$
Rate $\propto\left[\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br}\right]$
Similarly
Step II. $\mathrm{PhCD}_2 \stackrel{+}{\mathrm{C}_2} \mathrm{H}_2 \stackrel{\text { Fast }}{\longrightarrow} \mathrm{PhCD}=\mathrm{CHD}+\mathrm{H}^{+}$
Rate $\propto\left[\mathrm{PhCD}_2-\mathrm{CH}_2 \mathrm{Br}\right]$
Hence, $E_1$ reaction and first order kinetics.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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From JEE Advanced

Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamics process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by

T Δ X,

where

T

is temperature of the system and

Δ X

is the infinitesimal change in a thermodynamic quantity

X

of the system. For a mole of monatomic ideal gas

X = \frac{3}{2} R l n (\frac{T}{T_{A}}) + R l n (\frac{V}{V_{A}}) .

Here,

R

is gas constant,

V

is volume of gas,

T_{A}

and

V_{A}

are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

	List I		List II
$(a)$	Work done by the system in process $1 \to 2 \to 3$	$(p)$	$\frac{1}{3} R T_{0} l n 2$
$(b)$	Change in internal energy in process $1 \to 2 \to 3$	$(q)$	$\frac{1}{3} R T_{0}$
$(c)$	Heat absorbed by the system in process $1 \to 2 \to 3$	$(r)$	$R T_{0}$
$(d)$	Heat absorbed by the system in process $1 \to 2$	$(s)$	$\frac{4}{3} R T_{0}$
		$(t)$	$\frac{1}{3} R T_{0} (3 + l n 2)$
		$(u)$	$\frac{5}{6} R T_{0}$

If the process carried out on one mole of monatomic ideal gas is as shown in figure in the

T

V

-diagram with

P_{0} V_{0} = \frac{1}{3} R T_{0},

the correct match is,

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Column I	Column II
(A) \(CH _3- CHBr - CD _3\) on treatment with alc. KOH gives \(CH _2= CH - CD _3\) as a major product.	(P) E1 reaction
(B) \(Ph - CHBr - CH _3\) reacts faster than \(Ph - CHBr - CD _3\)	(Q) E2 reaction
(C) \(Ph - CH _2- CH _2 Br\) on treatment with \(C _2 H _5 OD / C _2 H _5 O ^{-}\)gives \(Ph - CD = CH _2\) as the major product	(R) El cb reaction
(D) \(PhCH _2 CH _2 Br\) and \(PhCH _2 CH _2 Br\) react with same rate	(S) First order reaction