Question

An alkane $(mol.wt.=86)$ on bromination gives only two monobromo derivatives (excluding stereoisomers). The alkane is:

• A. $C{H}_3-\underset{C{H}_3}{\underset{|}{C}}H-C{H}_2-C{H}_2-C{H}_3$
• B. $C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_2-C{H}_3$
• C. $C{H}_3-\underset{C{H}_3}{\underset{|}{C}}H-\underset{C{H}_3}{\underset{|}{C}}H-C{H}_3$
• D. $C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_3$

Answer 1

Answer: (C)

$C{H}_3-\underset{C{H}_3}{\underset{|}{C}}H-\underset{C{H}_3}{\underset{|}{C}}H-C{H}_3$

$C{H}_3-\underset{|}{CH}-C{H}_2-C{H}_2-C{H}_3\underset{h\nu }{\overset{B{r}_2}{\to }}5$ monobromoderivatives

$C{H}_3$ $A$

$C{H}_3$

$C{H}_3-\underset{|}{\stackrel{|}{C}}-C{H}_2-C{H}_3\underset{h\nu }{\overset{B{r}_2}{\to }}3$ monobromoderivatives

$C{H}_3$ $B$

$C{H}_3-\underset{|}{CH}-\underset{|}{CH}-C{H}_3\underset{h\nu }{\overset{B{r}_2}{\to }}2$ monobromoderivatives

$C{H}_3$ $C{H}_3$

Among them $A,B$ and $C$ have only molecular weight $86$. Now $A$ has $5$ different equivalent $H$ atoms and $B$ has $3$ different equivalent $H$ atoms, so $A$ and $B$ will give $5$ and $3$ monobromo derivatives respectively. But $C$ has $2$ different equivalent $H$ atoms, so $C$ will give $2$ monobromo derivatives.