Question

A hydrocarbon (Molar Mass = 72 g/mol) yields only one monochlorinated product and exactly two isomeric dichlorinated products upon photochemical chlorination. Which of the following is the correct structure of the hydrocarbon

• A. n-pentane
• B. isopentane
• C. neopentane

Answer 1

Answer: (C)

neopentane

Answer 2

The molar mass of the hydrocarbon is given as 72 g/mol. Let the molecular formula be $C_xH_y$.

$12x + y = 72$.

For an alkane $C_nH_{2n+2}$, the molar mass is $12n + (2n+2) = 14n + 2$.

Setting $14n + 2 = 72$, we get $14n = 70$, so $n = 5$.

The formula is $C_5H_{12}$. This is a pentane isomer.

There are three structural isomers of pentane:

1. n-pentane ($CH_3CH_2CH_2CH_2CH_3$)
2. isopentane (2-methylbutane, $CH_3CH(CH_3)CH_2CH_3$)
3. neopentane (2,2-dimethylpropane, $C(CH_3)_4$)

Photochemical chlorination involves the substitution of hydrogen atoms by chlorine atoms via a free radical mechanism. The number of possible monochlorinated products is equal to the number of distinct types of hydrogen atoms in the hydrocarbon molecule.

Let's analyze the types of hydrogen atoms in each isomer:

1. n-pentane ($CH_3^aCH_2^bCH_2^cCH_2^bCH_3^a$): There are three types of hydrogen atoms (a, b, c). Thus, n-pentane yields three monochlorinated products (1-chloropentane, 2-chloropentane, 3-chloropentane). This does not match the condition of yielding only one monochlorinated product.

2. isopentane ($CH_3^aCH^b(CH_3^c)CH_2^dCH_3^e$): Let's check symmetry. The two methyl groups attached to the branched carbon are equivalent ($CH_3^a$ and $CH_3^c$ are the same type).

$CH_3^a - CH^b(CH_3^a) - CH_2^d - CH_3^e$.

Types of hydrogens: primary on $C^1$ and $C^3$ (type a), tertiary on $C^2$ (type b), secondary on $C^4$ (type d), primary on $C^5$ (type e).

$CH_3^a$, $CH^b$, $CH_2^d$, $CH_3^e$. All four types are distinct.

Isopentane has 4 types of hydrogen atoms. Thus, it yields four monochlorinated products (1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 1-chloro-3-methylbutane, 3-chloro-2-methylbutane). This does not match the condition of yielding only one monochlorinated product.

3. neopentane ($C(CH_3)_4$):

$CH_3$ | $CH_3 - C - CH_3$ | $CH_3$

All four methyl groups attached to the central quaternary carbon are equivalent by symmetry. All 12 hydrogen atoms in neopentane are equivalent.

There is only one type of hydrogen atom. Thus, neopentane yields only one monochlorinated product: 1-chloro-2,2-dimethylpropane ($ClCH_2-C(CH_3)_3$). This matches the first condition.

Now let's check the dichlorinated products obtained from neopentane ($C_5H_{12}$). Dichlorination involves replacing two hydrogen atoms with two chlorine atoms. We need to find the number of distinct isomers of $C_5H_{10}Cl_2$.

In neopentane, all 12 hydrogens are equivalent. When we substitute two hydrogens, the two chlorine atoms can be on the same carbon atom or on different carbon atoms.

Let's consider the carbon framework: a central quaternary carbon bonded to four primary methyl carbons.

• Case 1: Both chlorine atoms are on the same carbon atom. Since all four methyl carbons are equivalent, we can place both Cl atoms on any one of them.

  Example: Replace two hydrogens on one methyl group ($CH_3$) to form $CHCl_2$.

  $CH_3$ | $CHCl_2 - C - CH_3$ | $CH_3$

  This is 1,1-dichloro-2,2-dimethylpropane.

• Case 2: The two chlorine atoms are on different carbon atoms. Since all four methyl carbons are equivalent, we can place one Cl on any methyl carbon and the second Cl on any other methyl carbon. As all methyl carbons are equivalent and bonded only to the central carbon, placing Cl on any two different methyl carbons will result in the same connectivity relative to the central carbon.

  Example: Replace one hydrogen on one methyl group ($CH_3$) to form $CH_2Cl$, and replace one hydrogen on another methyl group ($CH_3$) to form $CH_2Cl$.

  $CH_3$ | $CH_2Cl - C - CH_3$ | $CH_2Cl$

  This is 1,3-dichloro-2,2-dimethylpropane.

Are there any other distinct ways to place two chlorine atoms? No, the chlorine atoms must be on the methyl carbons as the central carbon has no hydrogens.

So, there are exactly two isomeric dichlorinated products formed from neopentane. This matches the second condition.

Therefore, the correct structure of the hydrocarbon is neopentane (2,2-dimethylpropane).