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?
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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.
Concept:
This problem involves two independent organic reactions. In the first part, an alkene showing geometrical (cis-trans) isomerism undergoes electrophilic addition of bromine. In the second part, a hydrocarbon reacts with \(NaNH_2\), indicating the presence of an acidic terminal alkyne hydrogen. The product formed on treatment with HBr must then be identified.
To solve the question, we identify compounds \(A\), \(B\), \(C\), and \(D\) step by step using the given information.
Step 1: Identify compound A
The molecular formula of A is
\[
C_4H_8
\]
and it exhibits cis-trans isomerism.
The isomers of \(C_4H_8\) are:
\[
CH_2=CHCH_2CH_3 \quad (\text{but-1-ene})
\]
\[
CH_3CH=CHCH_3 \quad (\text{but-2-ene})
\]
\[
(CH_3)_2C=CH_2 \quad (\text{2-methylpropene})
\]
Among these, only but-2-ene exhibits cis-trans isomerism because each carbon of the double bond is attached to two different groups.
Therefore,
\[
A = CH_3CH=CHCH_3
\]
i.e., but-2-ene.
Step 2: Determine compound B
Bromine in carbon tetrachloride adds across the double bond of an alkene.
\[
CH_3CH=CHCH_3
\xrightarrow{Br_2/CCl_4}
CH_3CHBrCHBrCH_3
\]
The product formed is
\[
2,3\text{-dibromobutane}
\]
Thus \(B\) must be the structure representing vicinal dibromide with bromine atoms attached to the second and third carbon atoms.
Among the given figures, this corresponds to the structure shown in
Fig. A.
Step 3: Identify compound C
The molecular formula is
\[
C_4H_6
\]
and it forms a sodium derivative with \(NaNH_2\).
Only terminal alkynes possess acidic hydrogen atoms that can be removed by \(NaNH_2\).
Among the isomers of \(C_4H_6\), the terminal alkyne is
\[
CH_3CH_2C \equiv CH
\]
which is but-1-yne.
Hence,
\[
C = CH_3CH_2C \equiv CH
\]
Step 4: Determine compound D
Addition of one mole of HBr to but-1-yne follows Markovnikov's rule.
Hydrogen adds to the terminal carbon and bromine adds to the more substituted carbon.
\[
CH_3CH_2C \equiv CH
\xrightarrow{HBr}
CH_3CH_2C(Br)=CH_2
\]
Thus D is
\[
2\text{-bromobut-1-ene}
\]
Among the given structures, this corresponds to the structure shown in
Fig. A.
Step 5: Select the correct option
We found that
\[
B = 2,3\text{-dibromobutane}
\]
and
\[
D = 2\text{-bromobut-1-ene}
\]
These two structures are represented together in
Fig. A.
Therefore the correct answer is
\[
\boxed{\text{Fig. A}}
\]
