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ANSWERS
Chapter 1: The Fundamentals
1. (B) Single and double bonds hold all the carbon atoms in this nanotube together; there-
fore, the answer is sp2 hybridization.
2. (B) Th e nitrogen atoms are triply bonded to each other. A triple bond involves sp 
hybridization.
3. (D) Carbon 1 is triply bonded (sp hybridization). Carbon 2 has a single bond to another 
carbon, two single bonds to hydrogen atoms, and one bond to a phosphorus atom. Carbon 
3 has a double bond to an oxygen atom and two single bonds, one to another carbon atom 
and one to a nitrogen atom.
4. (C) Beginning at the triple-bonded end, the bond is between two sp hybridized carbon 
atoms (carbons 1 and 2), between an sp and an sp3 hybridized carbon (carbons 2 and 3), 
and between two sp3 hybridized carbons (carbons 3 and 4); therefore, sp and sp3 hybridized 
orbitals are utilized.
5. (D) Th ere are fi ve on the ring containing the nitrogen, one on the methyl attached to 
the nitrogen, and two in the ethyl group, for a total of eight.
6. (A) All the nitrogen atoms indicated have a lone pair of electrons on the nitrogen. Nitro-
gen I is triple bonded and has a lone pair, indicating sp hybridization; nitrogen II has three 
single bonds and a lone pair, indicating sp3 hybridization; and nitrogen III has a double 
bond, a single bond, and a lone pair, indicating sp2 hybridization.
7. (A) Th e methyl cation has three groups around it, corresponding to sp2 hybridization, 
leaving an unhybridized p orbital to interact. Th e methyl anion would have four groups, 
three bonding pairs, and one nonbonding pair, corresponding to sp3 hybridization.
8. (A) You can eliminate answers B and D because of the charge separation, and answer A 
has the lowest formal charges on the atoms. Answer C features 10 electrons around carbon.
9. (D) Structure III is an ionic compound, not a resonance form. Th erefore, that eliminates 
answers A, B, and C.
10. (B) Th e presence of the highly electronegative fl uorine atoms withdraws electron den-
sity from the –O-H bond by an induction eff ect, weakening the bond and therefore increas-
ing the acidity. Th e more fl uorine atoms there are or the closer the fl uorine atoms are to 
the –OH, the greater the inductive eff ect and the stronger the acid.
11. (D) In ortho-nitrophenol there is resonance stabilization of the anion formed. Th ere 
is no comparable stabilization in the case of phenol.