An alkyl halide $\mathrm{C}_5 \mathrm{H}_{11}$ (A) reacts with ethanolic KOH to give an alkene ' $B$ ', which reacts with $\mathrm{Br}_2$ to give a compound ' $C$ ', which on dehydrobromination gives an alkyne ' $D$ '. On treatment with sodium metal in liquid ammonia, one mole of ' $D$ ' gives one mole of the sodium salt of ' $D$ ' and half a mole of hydrogen gas. Complete hydrogenation of ' $D$ ' yields a straight chain alkane. Identify $A, B, C$ and $D$. Give the reactions involved.
The reaction scheme involved in the problem is
Hydrogenation of alkyne $(D)$ gives straight chain alkane hence all the compounds $(A),(B)$, $(C)$ and (D) must be straight chain compounds. Alkyne (D) form sodium salt which proves that it is terminal alkyne. Involved reactions are as follows
It is important point that alkyl halide $(A)$ can not be 2-bromopentane because dehydrobromination of $(A)$ would have given 2-pentene as the major product in accordance with Markownikoff's rule.
896 mL vapour of a hydrocarbon ' $A$ ' having carbon $87.80 \%$ and hydrogen $12.19 \%$ weighs 3.28 g at STP. Hydrogenation of ' $A$ ' gives 2-methylpentane. Also ' $A$ ' on hydration in the presence of $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HgSO}_4$ gives a ketone ' $B$ ' having molecular formula $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}$. The ketone ' $B$ ' gives a positive iodoform test. Find the structure of ' $A$ ' and give the reactions involved.
To determine the molecular mass of hydrocarbon (A) 896 mL vapour of $\mathrm{C}_x \mathrm{H}_y(\mathrm{~A})$ weighs 3.28 g at STP
22700 mL vapour of $\mathrm{C}_x \mathrm{H}_y(A)$ weighs $\frac{3.28 \times 22700}{896} \mathrm{~g} / \mathrm{mol}$ at STP
$$=83.1 \mathrm{~g} / \mathrm{mol}$$
Hence, molecular mass of $C_x H_y(A)=83.1 \mathrm{~g} \mathrm{~mol}^{-1}$. To determine the empirical formula of hydrocarbon (A).
| Element | % | Atomic mass | Relative ratio | Relative no. of atoms | Simplest ratio |
|---|---|---|---|---|---|
| C | 87.8 | 12 | 7.31 | 1 | 3 |
| H | 12.19 | 1 | 12.19 | 1.66 | 4.98 $\approx 5$ |
Thus, Empirical formula of $A$ is $\mathrm{C}_3 \mathrm{H}_5$.
$\therefore$ Empirical formula mass $=36+5=41$.
$$n=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{83.1}{41}=2.02=2$$
Molecular mass is double of empirical formula mass.
$\therefore$ Molecular formula is $\mathrm{C}_6 \mathrm{H}_{10}$
To determine the structure of compounds $(A)$ and $(B)$
Hence, hydrogenation of hydrocarbon (A) requires 2 moles of hydrogen to form 2-methylpentane. Therefore, hydrocarbon(A) is an alkyne having five carbon atoms in a staight chain and a methyl substituent at position 2 . Thus, the possible structures for the alkyne $(A)$ are I and II.
Since, addition of $\mathrm{H}_2 \mathrm{O}$ to alkyne $(A)$ in presence of $\mathrm{Hg}^{2+}$, give a ketone which gives positive iodoform test, therefore, ketone $(B)$ must be a methyl ketone, i.e., it must contain a $\mathrm{COCH}_3$ group.
Now addition of $\mathrm{H}_2 \mathrm{O}$ to alkyne (II) should give a mixture of two ketones in which 2- methyl pentan -3 one (minor) and 4-methylpentan -2-one ketone (B) (which shows +ve iodoform test) predominates.
In contrast, addition of $\mathrm{H}_2 \mathrm{O}$ to alkyne (I) will give only one ketone, i.e., 4- methylpentan-2one which gives iodoform test.
Thus, hydrocabon $\mathrm{C}_x \mathrm{H}_y(A)$ is 4-methylpent -1-yne. 4- methylpentan -2 one (gives + ve iodoform test)
An unsaturated hydrocarbon ' $A$ ' adds two molecules of $\mathrm{H}_2$ and on reductive ozonolysis gives butane-1, 4-dial, ethanal and propanone. Give the structure of ' $A$ ', write its IUPAC name and explain the reactions involved.
The scheme of reaction is
In the presence of peroxide addition of HBr to propene takes place according to anti Markownikoff's rule but peroxide effect is not seen in the case of HCl and HI . Explain.
$\underset{\substack{\text { propene }}}{\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2}+\mathrm{HBr} \xrightarrow{\text { Peroxide }} \underset{\substack{n \text {-propyl bromide }}}{\mathrm{CH}_3-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}}$
The mechanism of the reaction is
Peroxide effect is effective only in the case of HBr and not seen in the case of HCl and HI . This is due to the following reasons.
(i) $\mathrm{H}-\mathrm{Cl}$ bond $(103 \mathrm{kcal} / \mathrm{mol})$ is stronger than $\mathrm{H}-\mathrm{Br}$ bond $(87 \mathrm{kcal} / \mathrm{mol})$ $\mathrm{H}-\mathrm{Cl}$ bond is not decomposed by the peroxide free radical whereas the $\mathrm{H}-\mathrm{I}$ bond is weaker ( $71 \mathrm{kcal} / \mathrm{mol}$ ) form iodine free radicals.
(ii) lodine free radical $\left(\mathrm{I}^{\circ}\right)$ formed as H - I bond is weaker but iodine free radicals readily combine with each other to form iodine molecules rather attacking the double bond.