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meriva
4 years ago
12

Write structures for the following compounds.(a) 3-ethyl-4-methylhexane (b) 3-ethyl-5-isobutyl-3-methylnonane(c) 4-tert-butyl-2-

methylheptane (d) 5-isopropyl-3,3,4-trimethyloctane

Chemistry
2 answers:
frutty [35]4 years ago
8 0

Answer:

Find attached answer to this question.

Explanation:

a. 4-Isopropyloctane has a chain of eight carbons, with an isopropyl group on the fourth carbon.

5-tert-Butyldecane has a chain of ten carbons, with a tert-butyl group on the fifth.

CH3

4-isopropyloctane

CHCH3 CH3

CH3 CH3CH2CH2CH2CH2CH2 CH

5-tert-butyldecane

CH3 CH3CH2CH2CH2CH2CH2CH2CH2 CH

CCH3 CH3

3-3 Nomenclature of Alkanes 93

di- means 2 tetra- means 4 hexa- means 6

tri- means 3 penta- means 5 hepta- means 7

b. Using this rule, we can construct names for some complicated structures. Let\u2019s fin-

ish naming the heptane on p. 90, shown here in the margin. This compound has an ethyl

group on C3 and three methyl groups on C2, C4, and C5. List the ethyl group alphabeti-

cally before the methyl groups, and give each of the four substituents a location number.

Problem-solving Hint

When substituents are alphabetized,

iso- is used as part of the alkyl group

name, but the hyphenated prefixes

are not. Thus, isobutyl is alphabetized

with i, but n-butyl, tert-butyl, and sec-

butyl are alphabetized with b. The

number prefixes di-, tri-, tetra-, etc.

are ignored in alphabetizing.

CH3 CH

2-bromobutane

Br

CH2CH3 CH3 CH

3-chloro-2-methylpentane

CH3 Cl

CH CH2CH3 CH3 CH

1,2-difluoropropane

F

CH2F

c. CH2CH(CH3)2CH2CH3

CH CHCH2 CH2 CH2CH3 CH3

S O LV E D P R O B L E M 3-2

Give a systematic (IUPAC) name for the following compound.

SOLUTION

The longest carbon chain contains eight carbon atoms, so this compound is named as an octane.

Numbering from left to right gives the first branch on C2; numbering from right to left gives

the first branch on C3, so we number from left to right.

CH CH2

CH3

CH CH CH2CH3

CH CH3

CH3

CH3

CH3

3-ethyl-2,4,5-trimethylheptane

7

65

4 3

2

1

CH3

CH3

CH3 CH3

CH3

CH3

CH3

CH

CH CH CH

CH2CH3

CH2

C

d. CH3 CH3

CH3

CH2 CH2 CH2 CH2 CH2

CH2

CH

CH3 CH3CH

CH

CH3 CH

Br CH2CH3

CH CH3(b)

CH3CH2

CH2CHCH3 CH3

lisabon 2012 [21]4 years ago
5 0

Answer:

Here's what I get  

Explanation:

(a) 3-Ethyl-4-methylhexane

The main chain is eight carbons long, with an ethyl group at C-3 and a methyl group at C-4

(b) 3-Ethyl-5-isobutyl-3-methylnonane

The main chain is nine carbons long, with an ethyl group and a methyl group on C-3 and an isobutyl group on C-5.

Note that the name isobutyl is acceptable in general use, but the preferred IUPAC name (PIN) for the group is now (since 2013) "2-methylpropan-2-yl".

The PIN for your compound is 3-ethyl-3-methyl-5-(2-methyl propan-2-yl)nonane.

(c) 4-<em>tert</em>-butyl-2-methylheptane

The main chain is seven carbons long, with a <em>tert</em>-butyl group at C-4 and a methyl group at C-2.

For some reason, the IUPAC has retained the name <em>tert</em>-butyl for the group, so the name you give is the PIN.

The prefix "<em>tert</em>" is written in italics.

(d) 5-isopropyl-3,3,4-trimethyloctane

The main chain is eight carbon atoms long, with two methyl groups on C-3 and one on C-4, and an isopropyl group on C-5.

The PIN for an isopropyl group is now "propan-2-yl", so the IUPAC name for your compound is 3,3,4-trimethyl-5-(propan-2-yl)octane.

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Whats the difference between a solution and a compound? (sorry if this is funny to ask)
Kipish [7]

Answer:

A compound is just pure  that is composed of elements that are chemically bonded in definite proportions

andd solution is umm well its a homogenous so basically saying

that it all just stays the same all the way through  

Explanation:

im not sure how else to explain it haah hope this helped  :P

3 0
3 years ago
In the following reaction, which species is reduced? Au(s) + 3NO3-(aq) + 6H+(aq) → Au3+(aq) + NO(g) + 3H2O (l) H+ N+5 O2- H2O Au
weqwewe [10]

Answer:

NO3-

Explanation:

Given the reaction equation;

Au(s) + 3NO3-(aq) + 6H+(aq)→Au3+(aq) + 3NO2(g) + 3H2O (l).

We can consider the oxidation states of species on the left and right hand sides of the reaction equation;

Au is in zero oxidation state on the left hand side and an oxidation state of +3 on the righthand side.

NO3- is in oxidation state of +5 on the righthand side and NO2 is in + 4 oxidation state.

H+ is in + 1 oxidation state on both the left and right hand sides of the reaction equation.

Since reduction has to do with a decrease in oxidation number, it follows that NO3- was reduced in the reaction.

6 0
3 years ago
The average speeds of gas molecules in cylinders A, B, C, and D are 0.01 m/s, 0.005 m/s, 0.1 m/s, and 0.5 m/s, respectively. Whi
Alik [6]
<span>We can relate the average speed of the gas molecules by its kinetic energy. The kinetic energy is the work done by the object due to its motion. It is represented by the formula of the half the velocity squared multiply by the mass of the object. The absolute zero theory is the assumption that at 0 Kelvin or -273.15 degrees Celsius, the gas molecules stop moving. It is because the temperature is severely cold enough to allow movement of the gas molecules. The gas molecule that is closest to the absolute zero is in cylinder B because it moves so slow and its movement is almost near to the absolute zero theory.</span>
4 0
3 years ago
If the solubility of a gas in water is 1.22g/2.75 atm, what is it’s solubility (in g/L) at 1.0 atm
Slav-nsk [51]

Answer : The solubility of a gas in water at 1 atm pressure is, 0.4436 g/L

Solution : Given,

Solubility of a gas in water = 1.22 g/L     (at 2.75 atm pressure)

At pressure = 1 atm,     Solubility of a gas = ?

Now we have to calculate the solubility of a gas.

At 2.75 atm pressure, the solubility of a gas in water = 1.22 g/L

At 1 atm pressure, the solubility of a gas in water = \frac{1atm}{2.75atm}\times 1.22g/L=0.4436g/L

Therefore, the solubility of a gas in water at 1 atm pressure is, 0.4436 g/L


5 0
3 years ago
The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com
Elanso [62]

Answer:

(a) Ksp=4.50x10^{-7}

(b) Ksp=1.55x10^{-6}

(c) Ksp=2.27x10^{-12}

(d) Ksp=1.05x10^{-22}

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)

Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L}   )^2\\\\Ksp=4.50x10^{-7}

(B) Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)

Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}

(C) NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)

Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}

(D) La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)

Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}

Best regards.

7 0
4 years ago
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