The alkane shown is monobrominated with Br2 and light. Select all correct statements.

Four constitutionally different monobromoalkanes are possible.

The tertiary bromide is the major product.

Four constitutionally different monobromoalkanes are possible.

The tertiary bromide is the major product.

Counting configurational stereoisomers as separate products, five monobromo products are possible.

All four constitutional monobromides are achiral.

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The alkane shown is monobrominated with $Br X_{2}$ and light. Select all correct statements.

AFour constitutionally different monobromoalkanes are possible.
BThe tertiary bromide is the major product.
CCounting configurational stereoisomers as separate products, five monobromo products are possible.
DAll four constitutional monobromides are achiral.

Answer: A, B

The compound shown is 2-methylbutane. Four constitutional bromides result (1-bromo-3-methylbutane, 2-bromo-2-methylbutane, 2-bromo-3-methylbutane, 1-bromo-2-methylbutane). Bromine is highly selective for the weakest, tertiary C-H bond, so 2-bromo-2-methylbutane is major. Two of the four have a stereocentre (2 enantiomers each), so stereoisomers give

1 + 1 + 2 + 2 = 6

products, not 5, and not all bromides are achiral.

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The alkane shown is monobrominated with BrX2​ and light. Select all correct statements.

Answer: A, B

More questions on Structure deduction and product counting from drawn compounds

The alkane shown is monobrominated with $Br X_{2}$ and light. Select all correct statements.