Q.65. Given below are four compounds: (a) n-propyl chloride (b) iso-propyl chloride (c) sec-butyl ch

Q: What is the correct answer for JEE Main Chemistry Q65?

The correct answer is A. Optical isomerism (chirality) requires a chiral carbon atom, which is a carbon atom bonded to four different groups. Let's examine each compound: (a) n-propyl chloride: CH$_3$-CH$_2$-CH$_2$-Cl. No chiral carbon. (b) iso-propyl chloride: CH$_3$-CH(Cl)-CH$_3$. No chiral carbon (the central carbon is bonded to two methyl groups). (c) sec-butyl chloride: CH$_3$-CH(Cl)-CH$_2$-CH$_3$. The second carbon

Q: Q.65. Given below are four compounds: (a) n-propyl chloride (b) iso-propyl chloride (c) sec-butyl chloride (d) neo-pentyl chloride Percentage of carbon in the one which exhibits optical isomerism is:

The correct answer is A. Optical isomerism (chirality) requires a chiral carbon atom, which is a carbon atom bonded to four different groups. Let's examine each compound: (a) n-propyl chloride: CH$_3$-CH$_2$-CH$_2$-Cl. No chiral carbon. (b) iso-propyl chloride: CH$_3$-CH(Cl)-CH$_3$. No chiral carbon (the central carbon is bonded to two methyl groups). (c) sec-butyl chloride: CH$_3$-CH(Cl)-CH$_2$-CH$_3$. The second carbon

Q: What are the options for this JEE Main Chemistry question?

A. 52 · B. 56 · C. 46 · D. 40

Question

Q.65. Given below are four compounds: (a) n-propyl chloride (b) iso-propyl chloride (c) sec-butyl chloride (d) neo-pentyl chloride Percentage of carbon in the one which exhibits optical isomerism is:

FirstInTest · Accepted Answer

Optical isomerism (chirality) requires a chiral carbon atom, which is a carbon atom bonded to four different groups. Let's examine each compound: (a) n-propyl chloride: CH$_3$-CH$_2$-CH$_2$-Cl. No chiral carbon. (b) iso-propyl chloride: CH$_3$-CH(Cl)-CH$_3$. No chiral carbon (the central carbon is bonded to two methyl groups). (c) sec-butyl chloride: CH$_3$-CH(Cl)-CH$_2$-CH$_3$. The second carbon atom (bonded to Cl) is attached to -H, -Cl, -CH$_3$, and -CH$_2$CH$_3$. These are four different groups. So, sec-butyl chloride is chiral and exhibits optical isomerism. (d) neo-pentyl chloride: (CH$_3$)$_3$C-CH$_2$-Cl. No chiral carbon. The carbon bonded to Cl is a primary carbon, and the adjacent quaternary carbon has three methyl groups. So, sec-butyl chloride (c) is the compound that exhibits optical isomerism. Now, we need to calculate the percentage of carbon in sec-butyl chloride. Formula: C$_4$H$_9$Cl. Molar mass of C$_4$H$_9$Cl = $(4 	imes 12) + (9 	imes 1) + (1 	imes 35.5) = 48 + 9 + 35.5 = 92.5$ g/mol. Mass of carbon in C$_4$H$_9$Cl = $4 	imes 12 = 48$ g. Percentage of carbon = $\frac{	ext{Mass of C}}{	ext{Molar mass of C}_4	ext{H}_9	ext{Cl}} 	imes 100 = \frac{48}{92.5} 	imes 100 \approx 51.89\%$. This is approximately 52%.

Answer

52

Answer

56

Answer

46

Answer

40

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