mcat AAMC full length 1
If 2-pentanol replaces 1-pentanol in the reaction shown in Figure 3,
the rate of is less
substitution
because:
A.
the C-O bond in 2-pentanol is stronger than the C-O bond in 1-
pentanol.
B
.there is a competing elimination reaction that slows the rate of
substitution.
C.
there is more steric hindrance at the oxygen atom in 2-pentanol than in
1-pentanol,
making protonation less
likely.
D
.there is more steric hindrance at the 2-position of 2-pentanol than at the 1-
position
pentano of 1-
l.
figure 3: Is an Sn2 reaction correct answersissue:
content/reasoning
Correct answer is D.Sn2 is preferred if there is less steric hindrance- less
substituted
carbon is
preferred
you thought the answer was c. but steric hindrance has to do with the
carbon
case. in this
there will be no difference in SH for the OXYGEN atom whether it
is at theor1 2 carbon. however there will be a difference in the SH for
carbon
the carbons
themselve
s!
if a solution containing the compounds shown in Figure 4, is injected into a
gas-liquid
chromatograph, the first peak observed in the gc trace is
attributable to which
compound
?
A.2-Methyl-2-
butanol
B.2-Methyl-2-
butene
C.2-Chloro-2-
methylbutane
D.2-Bromo-2-
methylbutaneReaction:
2-Methyl-2-butanol -->
into
2-Chloro-2-methylbutane
2-Bromo-2-methylbutane
(40%)
(40%)
2-Methyl-2-butene (20%) correct
answerscorrect
In gas-liquid chromatography, the first peak to emerge will be from the least
polar, most volatile compound. 2-Methyl-2-butene is nonpolar and the most
volatile because it only has London forces for intermolecular attraction.
WEAKER COMPOUND WILL BECOME GAS FIRST. stationary phase allows polar
molecules to elute the SLOWEST, therefore
,nonpolar molecules elute the quickest. also has to do with MW, larger
MW will elute
slower
too.
Gas chromatography correct answersgas flowing through a coated tube
separates
compounds by their size, weight, and chemical reactivity with the coating of
the tube or things up by boiling pointboiling point related to IMF (stronger
columnsplits
IMF will have
higher
BP)
What is the product of the reaction of Compound 1 (shown below) with
HBr by the
pathway shown in Figure
3?
Compound
1
A.
(R)-1-bromo-1-
deuteriohexane
B
.(S)-1-bromo-1-
deuteriohexane
C.
(S)-1-bromo-1-
deuteriopentane
D
.(R)-1-bromo-1-deuteriopentane correct answerswrong:
content
compound will undergo an sn2 reaction. sn2 has inversion of the
sterocenter.isotopes
hydeogron priority is
of as follows : tritum > deuterium > normal hydrogen.
but still lower
priority than anything else. since original compound is in R config (1-2-3
clockwise),
inversion (which happens in a sn2 rxn) will caiuse it to switch to opp
config so S!
answer is
D
sn1 vs sn2 correct answerssn2: complete inversion, since LG has to be opp
side of it
group, new
will alwsy switch from R to S or
S to R
sn1: when carbocation forms, it is an sp2 planar molecule. this means new
group can
attack from front or back. meaning we get a racemic mixture
of S and R
speed of sound = v = wavelength
times
so howf do we change speed of sound? correct answerschanging
wavelengthWILL
frequency or NOT change speed of sound. only way to change speed of
sound isthe
change to medium it is passing
through
stiffer medium = faster sound waves (stronger IMFS so any
distrubance quicker
transmitter will be (strong connections quicker
communication))
density = more dense = slower sound waves (more mass = more
intertia =to
resistant more
changes = slower at transmitting those
distrubances)
, stiffness >
density
speed of sound = solids > liquids >
gases
density and temp correct answershotter air =
less dense
colder air = more
dense
open open standing wave (closed closed) correct answersanti nodes at
both end. =
wavelength
2l/n
simplest is that one length is half a
wavelength
open closed standing wave correct answersanti node one end node
other end. =
wavelength
4l/n
simplest is that one length is a quarter of a
wavelength
Which of the following will decrease the percentage ionization of 1.0 M
acetic acid,
CH3CO2H(aq)
?
A.
Chlorinating the CH3
group
B
.Diluting the
solution
C.
Adding concentrated
HCl(aq)
D
.Adding a drop of basic indicator correct answerscorrect
answer c
HCl is a strong acid that will increase the amount of H+ in solution and
thuspercentage
the decrease of CH3CO2H that
ionizes.
ionization means taking a neutral atom and making it charged. loss or gain
of chargeH
through
atoms.
For this problem, you are trying to decrease the amount of ionization of
acetic
you acidwant
don't (AKAit to release its H+). In your chemical reaction, it's fine but
by diluting
solution the water) you will be promoting the reaction to go forward
(adding
(increasing
ionization
).
The reason adding HCl is correct is because it is a strong acid so it .
dissociates
That means100%
a lot of H+ are being added to the solution, causing your
reaction to (decreasing
backwards go
ionization).
CH3CO2H == CH3CO2- +
H+
The intensity of the radiation emitted by the oxygen sensor is directly
proportional to the:
A.
propagation speed of the
radiation.
B.
wavelength of the
radiation.
If 2-pentanol replaces 1-pentanol in the reaction shown in Figure 3,
the rate of is less
substitution
because:
A.
the C-O bond in 2-pentanol is stronger than the C-O bond in 1-
pentanol.
B
.there is a competing elimination reaction that slows the rate of
substitution.
C.
there is more steric hindrance at the oxygen atom in 2-pentanol than in
1-pentanol,
making protonation less
likely.
D
.there is more steric hindrance at the 2-position of 2-pentanol than at the 1-
position
pentano of 1-
l.
figure 3: Is an Sn2 reaction correct answersissue:
content/reasoning
Correct answer is D.Sn2 is preferred if there is less steric hindrance- less
substituted
carbon is
preferred
you thought the answer was c. but steric hindrance has to do with the
carbon
case. in this
there will be no difference in SH for the OXYGEN atom whether it
is at theor1 2 carbon. however there will be a difference in the SH for
carbon
the carbons
themselve
s!
if a solution containing the compounds shown in Figure 4, is injected into a
gas-liquid
chromatograph, the first peak observed in the gc trace is
attributable to which
compound
?
A.2-Methyl-2-
butanol
B.2-Methyl-2-
butene
C.2-Chloro-2-
methylbutane
D.2-Bromo-2-
methylbutaneReaction:
2-Methyl-2-butanol -->
into
2-Chloro-2-methylbutane
2-Bromo-2-methylbutane
(40%)
(40%)
2-Methyl-2-butene (20%) correct
answerscorrect
In gas-liquid chromatography, the first peak to emerge will be from the least
polar, most volatile compound. 2-Methyl-2-butene is nonpolar and the most
volatile because it only has London forces for intermolecular attraction.
WEAKER COMPOUND WILL BECOME GAS FIRST. stationary phase allows polar
molecules to elute the SLOWEST, therefore
,nonpolar molecules elute the quickest. also has to do with MW, larger
MW will elute
slower
too.
Gas chromatography correct answersgas flowing through a coated tube
separates
compounds by their size, weight, and chemical reactivity with the coating of
the tube or things up by boiling pointboiling point related to IMF (stronger
columnsplits
IMF will have
higher
BP)
What is the product of the reaction of Compound 1 (shown below) with
HBr by the
pathway shown in Figure
3?
Compound
1
A.
(R)-1-bromo-1-
deuteriohexane
B
.(S)-1-bromo-1-
deuteriohexane
C.
(S)-1-bromo-1-
deuteriopentane
D
.(R)-1-bromo-1-deuteriopentane correct answerswrong:
content
compound will undergo an sn2 reaction. sn2 has inversion of the
sterocenter.isotopes
hydeogron priority is
of as follows : tritum > deuterium > normal hydrogen.
but still lower
priority than anything else. since original compound is in R config (1-2-3
clockwise),
inversion (which happens in a sn2 rxn) will caiuse it to switch to opp
config so S!
answer is
D
sn1 vs sn2 correct answerssn2: complete inversion, since LG has to be opp
side of it
group, new
will alwsy switch from R to S or
S to R
sn1: when carbocation forms, it is an sp2 planar molecule. this means new
group can
attack from front or back. meaning we get a racemic mixture
of S and R
speed of sound = v = wavelength
times
so howf do we change speed of sound? correct answerschanging
wavelengthWILL
frequency or NOT change speed of sound. only way to change speed of
sound isthe
change to medium it is passing
through
stiffer medium = faster sound waves (stronger IMFS so any
distrubance quicker
transmitter will be (strong connections quicker
communication))
density = more dense = slower sound waves (more mass = more
intertia =to
resistant more
changes = slower at transmitting those
distrubances)
, stiffness >
density
speed of sound = solids > liquids >
gases
density and temp correct answershotter air =
less dense
colder air = more
dense
open open standing wave (closed closed) correct answersanti nodes at
both end. =
wavelength
2l/n
simplest is that one length is half a
wavelength
open closed standing wave correct answersanti node one end node
other end. =
wavelength
4l/n
simplest is that one length is a quarter of a
wavelength
Which of the following will decrease the percentage ionization of 1.0 M
acetic acid,
CH3CO2H(aq)
?
A.
Chlorinating the CH3
group
B
.Diluting the
solution
C.
Adding concentrated
HCl(aq)
D
.Adding a drop of basic indicator correct answerscorrect
answer c
HCl is a strong acid that will increase the amount of H+ in solution and
thuspercentage
the decrease of CH3CO2H that
ionizes.
ionization means taking a neutral atom and making it charged. loss or gain
of chargeH
through
atoms.
For this problem, you are trying to decrease the amount of ionization of
acetic
you acidwant
don't (AKAit to release its H+). In your chemical reaction, it's fine but
by diluting
solution the water) you will be promoting the reaction to go forward
(adding
(increasing
ionization
).
The reason adding HCl is correct is because it is a strong acid so it .
dissociates
That means100%
a lot of H+ are being added to the solution, causing your
reaction to (decreasing
backwards go
ionization).
CH3CO2H == CH3CO2- +
H+
The intensity of the radiation emitted by the oxygen sensor is directly
proportional to the:
A.
propagation speed of the
radiation.
B.
wavelength of the
radiation.
