5 Steps to a 5 AP Chemistry 2019

(Marvins-Underground-K-12) #1

278 ❯ STEP 4. Review the Knowledge You Need to Score High


C

C N

CH 3 H

CH 3 CH 3

CH

HC CH

HC

C
H


  1. The organic compound shown above would be
    classified as:
    (A) an organic base
    (B) an ether
    (C) an alcohol
    (D) an aldehyde

  2. A carboxylic acid may be represented as:


(A) ROH
(B) RCHO
(C) R–O–R′
(D) RCOOH


  1. Which basic concept from the first half of the
    textbook best explains why carbon in stable
    organic compounds has four bonds?
    (A) VSEPR
    (B) octet rule
    (C) Avogadro’s hypothesis
    (D) electronegativity

  2. Which of the following is the one reaction that
    all organic compounds will undergo?
    (A) neutralization
    (B) precipitation
    (C) combustion
    (D) polymerization


❯ Answers and Explanations



  1. C—The general formula CnH2n means that
    1 mole of H 2 O will form per mole of empirical
    formula unit, regardless of the value of n. The
    moles of water formed are the mass of the alkene
    divided by the empirical formula mass:

    














(14 g alkene)
1 mol alkene
14 g alkene

1 mol HO
1 mol alkene

(^2)
= 1.0 mole



  1. C—Organic bases are, in general, amines (con-
    tain N). An ether would have an oxygen atom
    single-bonded to two carbon atoms (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; how-
    ever, in alcohols, the hydrogen atom in the
    general formula (ROH) is attached to an oxygen
    atom, whereas in aldehydes, the hydrogen atom
    in the general formula (RCHO) is attached to
    a carbon atom. Ketones have oxygen double-
    bonded to a carbon not at the end of a chain.
    3. D—The combination of an OH group attached
    to a carbon atom and an oxygen atom with a
    double bond to the same carbon atom is charac-
    teristic of a carboxylic acid.
    4. A—Redrawing the structures may help you to
    recognize the correct answer. An optical isomer
    must be a carbon atom with four different groups
    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 like A, but two of
    the groups, the –CH 2 CH 3 groups, are the same.
    Many biologically important molecules are
    optical isomers.

  2. D—A is an alcohol, B is an aldehyde, and C is
    an ether.

  3. D—The highest pH is the result of dissolving a
    base in water. A is an aldehyde, which is neutral.
    B is an alcohol, which is neutral (or very slightly
    acidic). C is a carboxylic acid, which is an acid.
    D is an amine, which is a base.

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