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tekilochka [14]
3 years ago
13

(a) How many stereoisomers are possible for 4-methyl-1,2-cyclohexanediol? ___ (b) Name the stereoisomers formed by oxidation of

(S)-4-methylcyclohexene with osmium tetroxide. If there is only one stereoisomer formed, leave the second space blank. Isomer #1: Isomer #2: (c) Is the product formed in step (b) optically active? _____

Chemistry
1 answer:
nataly862011 [7]3 years ago
6 0

Answer:

See explanation

Explanation:

For the first part of the question, we have to check the <u>chiral carbons</u> in 4-methyl-1,2-cyclohexanediol. In this case carbons, <u>1 and 2 are chiral</u>, if we have 2 chiral carbons we will have 4 isomers. We have to remember that formula 2^n in which "n" is the number of chiral carbons, so:

<u>2^n = 2^2 = 4 isomers</u>

<u>And the isomers that we can have are:</u>

1) (1R,2S)-4-methylcyclohexane-1,2-diol

2) (1S,2S)-4-methylcyclohexane-1,2-diol

3) (1S,2S)-4-methylcyclohexane-1,2-diol

4) (1S,2R)-4-methylcyclohexane-1,2-diol

See figure 1

For the second part of the question, we have to remember that the oxidation with OsO_4 is a <u>syn addition</u>. In other words, the "OHs" are added in the <u>same plane</u>. In this case, we have the methyl group with a wedge bond, so the "OH" groups will have a dashed bond due to the <u>steric hindrance</u>. Due to this we only can have <u>1 isomer</u> ((1S,2R,4S)-4-methylcyclohexane-1,2-diol). Finally, on this molecule, <u>we dont have any symmetry planes</u> (this characteristic will cancel out the optical activity), so <u>the product of this reaction has optical activity</u>.

See figure 2

I hope it helps!

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If 1.3 L of C3H8 combusts according to the equation below, how much CO2 will be produced?
SCORPION-xisa [38]

Answer:

0.162 moles of CO₂ are produced by this reaction

Explanation:

The reaction is:

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As we have the volume of propane, we need to know the mass that has reacted, so we apply density's concept.

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0.00183 g/mL . 1300 mL = Mass → 2.379 g

We determine the moles → 2.379 g . 1mol / 44 g = 0.054 moles

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6 0
3 years ago
A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati
prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

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k = rate constant

t = time passed by the sample = 48.0 hr

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a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

k=\frac{2.303}{48.0}\log\frac{53500}{42520}

k=9.98\times 10^{-2}hr^{-1}

Now we have to calculate the half-life.

k=\frac{0.693}{t_{1/2}}

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t_{1/2}=6.94hr

Therefore, the half life of 28-Mg in hours is, 6.94

8 0
3 years ago
5 Questions to answer please
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Suppose an ice cube weighing 36.0 g at a temperature of 10°C is placed in 360 g water at a temperature of 20°C. Calculate the te
Scilla [17]

Answer:

10.44 °C

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4 0
3 years ago
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