1.
Which radical is most stable?
Correct Answer
E. CH2=CH-ĊH2
Explanation
An allyl radical is the most stable (even more stable than a 3° radical) because the "true" structure of the allyl radical is a hybrid of two resonance structures. The π bond and the unpaired electron are delocalized over the entire molecule which helps to stabilize the allyl radical. (pg 551-552)
2.
What is the geometry of a carbon radical?
Correct Answer
C. Trigonal planar
Explanation
A carbon radical is trigonal planar, with 120° bond angles. (pg. 537)
3.
What is the orbital hybridization of a carbon radical?
Correct Answer
C. Sp²
Explanation
A carbon radical is sp² hybridized, similar to carbocations. The unhybridized p orbital contains the unpaired electron and extends above and below the trigonal planar carbon. (pg. 537)
4.
Fill in the blank correctly:The _______ the bond dissociation energy for a C-H bond, the more stable the resulting carbon radical.
Correct Answer
A. Lower
Explanation
Cleavage of the stronger 1° C-H bond to form a 1° carbon radical requires more energy than cleavage of the weaker 2° C-H bond to form a 2° carbon radical. This makes the 2° radical more stable because less energy is required for its formation. Cleavage of the WEAKER bond forms the MORE stable radical. (pg. 538)
5.
Alkanes react with halogens in the presence of __________ to form alkyl halides.
Correct Answer(s)
A. Light or heat
B. Hv or Δ
Explanation
In the presence of light (hv) or heat (Δ), alkanes react with halogens to form alkyl halides. The energy from light or heat is required to form the initial halide radical. (pg. 540)
6.
What are the important differences between chlorination and bromination?
Correct Answer(s)
A. Chlorination is faster than bromination.
C. Chlorination yields a mixture of products, while bromination usually yields only one major product.
Explanation
Chlorination is faster than bromination; its rate-determining step is exothermic. Chlorination also yields a mixture of products because the rate-determining step is exothermic, and so all radicals are formed, leading to a mixture of products. Bromination has a slow, endothermic rate-determining step, and so it favors the formation of the most stable radical, leading to a single major product. (pg. 544-547)
7.
What is the reactivity-selectivity principle?
Correct Answer
A. Less reactive reagents are more selective.
Explanation
Less reactive reagents are more selective. In bromination (which is less reactive than chlorination), the major (and sometimes only) product results from cleavage of the WEAKEST C-H bond. (pg. 545)
8.
Describe the 2-step method for converting an alkane to an alkene.
Correct Answer
B. Radical halogenation with Br2 followed by elimination of HBr.
Explanation
Halogenation is a useful tool because it adds a functional group to an otherwise unreactive alkane, making an alkyl halide. This alkyl halide can then be converted to an alkene by elimination, or to alcohols or ethers by nucleophilic substitution. (pg. 547)
9.
What is/are the monohalogenation product(s)?n-butane + chlorine + hv --> ______________
Correct Answer(s)
A. 1-chlorobutane
B. (R)-2-chlorobutane
C. (S)-2-chlorobutane
Explanation
Monochlorination of n-butane forms two constitutional isomers by replacement of either a 1° or 2° hydrogen. Radical abstraction of the 2° hydrogen forms a planar radical that can react with Cl2 from either side, resulting in a racemic mixture of two enantiomers. (pg. 548)
10.
A radical halogenation reaction at a stereogenic center results in ________________ .
Correct Answer
B. A racemic mixture of enantiomers
Explanation
Abstraction of a hydrogen at a stereogenic center forms a trigonal planar sp² hybridized radical that is now achiral. This achiral radical then reacts with Cl2 from either side to form a new stereogenic center, resulting in an equal amount of two enantiomers -- a racemic mixture. (pg. 549)