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Margaret [11]
3 years ago
7

What classification of alcohol would be synthesized from the reaction of 2-butanone with a Grignard reagent? From the reaction o

f butanal with a Grignard reagent?A. A secondary alcohol; a secondary alcoholB. A tertiary alcohol; a secondary alcoholC. A secondary alcohol; a primary alcoholD. A tertiary alcohol; a primary alcohol
Chemistry
2 answers:
r-ruslan [8.4K]3 years ago
8 0

Answer:

B. A tertiary alcohol; a secondary alcohol

Explanation:

Grignard reagent has a formula RMgX where X is a halogen, and R is an alkyl or aryl group.

Grignard reagent reacts with aldehydes and ketones to form alcohol. However the classification (that is; Primary, secondary and tertiary)  varies depending on which it reacts with.

2-butanone is a ketone.

Ketones reacts with Grignard reagent to form tertiary alcohol

Butanal is an aldehyde.

Aldeyde reacts with grignard reagent to form Secondary alcohol

The answer is; Tertiary alcohol; Secondary alcohol

mart [117]3 years ago
7 0

Answer:

It takes 30 minutes to feel the effects of alcohol.

Explanation:

It may take an hour to metabolize a drink, but it takes approximately thirty minutes before you feel alcohol's effects. This is a good gauge for pacing yourself. Drinking more than one drink every 30 minutes means you are probably drinking too much, too fast.

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The vapor pressure of pure pentane and pure hexane are 425 torr and 151 torr respectively. equal moles of pentane and hexane are
nadezda [96]

The mole fraction of pentane in the vapor is 0.291

Vapour pressure rises with temperature and is a measurement of a substance's propensity to transform into a gaseous or vapour state. The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapour pressure present at the liquid's surface.

The number of moles of a particular component in the solution divided by the overall number of moles in the sample solution is known as the mole fraction.

Using the formula for vapour pressure,

vapour pressure = P^{}°_{hex} × X_{hex} + P^{}°_{pen} × X_{pen}

vapour pressure = 151 ×  (1-X_{pen}) + 425 × X_{pen}

240 = 151 - 151X_{pen} + 425X_{pen}

240 - 151 = -  151X_{pen} + 425 X_{pen}

89 = 274 X_{pen}

\frac{89}{274} = X_{pen}

0.291 = X_{pen}

Therefore,  the mole fraction of pentane in the vapor is 0.291

Learn more about vapour pressure here;

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