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anastassius [24]
3 years ago
6

Provide a stepwise synthesis of 1-cyclopentylethanamine using the Gabriel synthesis. Collapse question part Testbank, Question 0

65a Correct answer. Your answer is correct. Using the reagents below, list in order (by letter, no period) those necessary to provide this synthesis. a. Mg(OH)2 b. KOH c. 1-bromo-1-cyclohexylethane d. 1-bromo-1-cyclopentylethane e. NH3 f. xs LiAlH4 g. H3O , then OH- Step 1 Step 2 Step 3 Entry field with correct answer Entry field with correct answer Entry field with correct answer
Chemistry
1 answer:
ra1l [238]3 years ago
8 0

Answer:

Azide synthesis is the first method on the table of synthesis of primary amines. The Lewis structure of the azide ion, N3−, is as shown below.

an azide ion

An “imide” is a compound in which an N−−H group is attached to two carbonyl groups; that is,

imide linkage

You should note the commonly used trivial names of the following compounds.

phthalic acid, phthalic anhydride, and phthalimide

The phthalimide alkylation mentioned in the reading is also known as the Gabriel synthesis.

If necessary, review the reduction of nitriles (Section 20.7) and the reduction of amides (Section 21.7).

Before you read the section on reductive amination you may wish to remind yourself of the structure of an imine (see Section 19.8).

The Hofmann rearrangement is usually called the Hofmann degradation. In a true rearrangement reaction, no atoms are lost or gained; however, in this particular reaction one atom of carbon and one atom of oxygen are lost from the amide starting material, thus the term “rearrangement” is not really appropriate. There is a rearrangement step in the overall degradation process, however: this is the step in which the alkyl group of the acyl nitrene migrates from carbon to nitrogen to produce an isocyanate.

Explanation:

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6 0
3 years ago
Please help, I really don’t understand this!!!
kap26 [50]

<u>Analysing the Question:</u>

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: <em>for every 1 mole of Glucose, we need 6 moles of Oxygen</em>

<u>Moles of Glucose used in the reaction:</u>

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

<u>Mass of Oxygen required:</u>

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen  = 32 / 30

Mass of Oxygen = 1.06 grams

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