Question

An isomer of \(C_4H_8\) (A) exhibits cis-trans isomerism. Reaction of A with \(Br_2/CCl_4\) gave B. Another organic compound \(C_4H_6\) (C) forms sodium derivative with \(NaNH_2\). Reaction of C with HBr gave D. What are B and D respectively? center

center Options :

• A. Fig. A
• B. Fig. B
• C. Fig. C
• D. Fig. D

Hint:

For a hydrocarbon to exhibit cis-trans isomerism, each carbon of the double bond must be attached to two different groups. Also remember that terminal alkynes react with \(NaNH_2\) because of their acidic hydrogen atom. Addition of HBr to terminal alkynes generally follows Markovnikov orientation in the absence of peroxide.

Answer 1

The Correct Option is A

Step 1: Read what is given.
Compound A is $C_4H_8$ and shows cis-trans isomerism. A reacts with $Br_2/CCl_4$ to give B. Separately, C is $C_4H_6$ and forms a sodium derivative with $NaNH_2$. C reacts with HBr to give D. We find B and D.
Step 2: Identify A.
Among the $C_4H_8$ alkenes, only but-2-ene, $CH_3CH=CHCH_3$, shows cis-trans isomerism, because each double-bond carbon holds two different groups. So A is but-2-ene.
Step 3: Find B.
Bromine in $CCl_4$ adds across the double bond: \[ CH_3CH=CHCH_3 \xrightarrow{Br_2/CCl_4} CH_3CHBrCHBrCH_3. \] So B is 2,3-dibromobutane (a vicinal dibromide).
Step 4: Identify C.
C is $C_4H_6$ and reacts with $NaNH_2$, which only removes the acidic hydrogen of a terminal alkyne. So C is but-1-yne, $CH_3CH_2C\equiv CH$.
Step 5: Find D.
Adding one HBr to but-1-yne follows Markovnikov's rule, placing Br on the more substituted carbon. This gives a bromo-alkene, the structure shown for D.
Step 6: Match the figure.
The figure that shows both the 2,3-dibromobutane (B) and the Markovnikov bromo-alkene (D) is the correct one. \[ \boxed{\text{Fig. A}} \]