Organic Chemistry 275
Answers and Explanations
- Cycloalkanes are hydrocarbons with the general
formula CnH 2 n. If a 0.500 g sample of any alkene
is combusted in excess oxygen, how many moles
of water will form?
(A) 0.50
(B) 0.072
(C) 0.036
(D) 1.0
(E) 0.018
2.
The organic compound shown above would be
classified as
(A) an organic base
(B) an ether
(C) an alcohol
(D) an aldehyde
(E) a ketone
- Which of the following compounds is optically
active?
(A) CH 3 CHClCH 2 CH 2 CH 3
(B) CH 3 CH=CHCH 2 CH 3
(C) CH 3 CH 2 CHClCH 2 CH 3
(D) CH 3 CH 2 CH 2 CH 2 OH
(E) CH 3 CH 2 CH 2 CH 2 CH 3 - A carboxylic acid may be represented as:
(A) ROH
(B) RCHO
(C) R-O-R′
(D) RCOOH
(E) RCOOR′
O–H
|
CH 3 –CH–CH 2 –CH 3
Review Questions
You have 5 minutes. You may not use a calculator. You may use the periodic table at the
back of the book. For each question, circle the letter of your choice.
- C—The general formula, CnH 2 n, means that
1 mol of H 2 O will form per mole of empirical
formula unit, regardless of the value of n. The
moles of water formed is the mass of the alkene
divided by the empirical formula mass.
(0.500 g alkene)(l mol alkene/14 g alkene)(l mol
H 2 O/mol alkene)
=0.036 mol
- C—Organic bases are, in general, amines
(contain N). An ether would have an oxygen
single-bonded to two carbons (R groups). An
aldehyde has oxygen double-bonded to a carbon
at the end of a chain. Aldehydes (RCHO) and
alcohols (ROH) are often confused, because of
the similarity in their general formulas. Ketones
have oxygen double-bonded to a carbon not at
the end of a chain.
- A—Redrawing the structures may help you to
recognize the correct answer. An optical isomer
must be a carbon atom with four differentgroups
attached to it. For A the groups on the second
carbon are: CH 3 −,H, Cl, and −CH 2 CH 2 CH 3.
Answer C is misleading. It is similar to A, but two
of the groups, the −CH 2 CH 3 groups, are the
same. - D—A=alcohol B =aldehyde C =ether E =ester