IIT-JEE 2010 · Paper 1 · Chemistry Q11

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ChemistryMultiple correct+3Question 11 of 84

In the Newman projection for 2,2-dimethylbutane, X and Y can respectively be

Newman projection along a C-C bond of 2,2-dimethylbutane: the front carbon carries X (up), H (lower left) and H (lower right); the back carbon carries CH₃ (upper left), CH₃ (upper right) and Y (down).

H and H

H and

\ce{C2H5}

\ce{C2H5}

and H

\ce{CH3}

and

\ce{CH3}

Answer and worked solution

Official answer: B, D

2,2-Dimethylbutane is $\ce{CH3-C(CH3)2-CH2-CH3}$ (C1-C4). The back carbon in the projection carries two $\ce{CH3}$ groups and Y; the front carbon carries two H atoms and X.

Along C1-C2: front C1 has H, H, H; back C2 has $\ce{CH3}$ , $\ce{CH3}$ , $\ce{C2H5}$ . So X = H, Y = $\ce{C2H5}$ : (B).
Along C3-C2: front C3 has H, H, $\ce{CH3}$ (C4); back C2 has three $\ce{CH3}$ . So X = $\ce{CH3}$ , Y = $\ce{CH3}$ : (D).

(A) would need a back carbon with two methyls and an H, and (C) a front carbon bearing ethyl and two H with the back carbon bearing H; neither exists in this molecule.

(B), (D)

Topics:Conformations Alkanes

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IIT-JEE 2010 · Paper 1, Chemistry question 11