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10-chloro-1-decanol

Input interpretation

10-chloro-1-decanol
10-chloro-1-decanol

Chemical names and formulas

formula | Cl(CH_2)_10OH Hill formula | C_10H_21ClO name | 10-chloro-1-decanol IUPAC name | 10-chlorodecan-1-ol alternate names | 10-chlorodecan-1-ol mass fractions | C (carbon) 62.3% | Cl (chlorine) 18.4% | H (hydrogen) 11% | O (oxygen) 8.3%
formula | Cl(CH_2)_10OH Hill formula | C_10H_21ClO name | 10-chloro-1-decanol IUPAC name | 10-chlorodecan-1-ol alternate names | 10-chlorodecan-1-ol mass fractions | C (carbon) 62.3% | Cl (chlorine) 18.4% | H (hydrogen) 11% | O (oxygen) 8.3%

Lewis structure

Draw the Lewis structure of 10-chloro-1-decanol. Start by drawing the overall structure of the molecule:  Count the total valence electrons of the carbon (n_C, val = 4), chlorine (n_Cl, val = 7), hydrogen (n_H, val = 1), and oxygen (n_O, val = 6) atoms: 10 n_C, val + n_Cl, val + 21 n_H, val + n_O, val = 74 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), chlorine (n_Cl, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): 10 n_C, full + n_Cl, full + 21 n_H, full + n_O, full = 138 Subtracting these two numbers shows that 138 - 74 = 64 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 32 bonds and hence 64 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 74 - 64 = 10 electrons left to draw: Answer: |   |
Draw the Lewis structure of 10-chloro-1-decanol. Start by drawing the overall structure of the molecule: Count the total valence electrons of the carbon (n_C, val = 4), chlorine (n_Cl, val = 7), hydrogen (n_H, val = 1), and oxygen (n_O, val = 6) atoms: 10 n_C, val + n_Cl, val + 21 n_H, val + n_O, val = 74 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), chlorine (n_Cl, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): 10 n_C, full + n_Cl, full + 21 n_H, full + n_O, full = 138 Subtracting these two numbers shows that 138 - 74 = 64 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 32 bonds and hence 64 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 74 - 64 = 10 electrons left to draw: Answer: | |

3D structure

3D structure
3D structure

Basic properties

molar mass | 192.7 g/mol phase | liquid (at STP) melting point | 12.5 °C boiling point | 127 °C (measured at 266.6 Pa) density | 0.952 g/cm^3
molar mass | 192.7 g/mol phase | liquid (at STP) melting point | 12.5 °C boiling point | 127 °C (measured at 266.6 Pa) density | 0.952 g/cm^3

Units

Liquid properties (at STP)

density | 0.952 g/cm^3 refractive index | 1.46
density | 0.952 g/cm^3 refractive index | 1.46

Units

Chemical identifiers

CAS number | 51309-10-5 PubChem CID number | 123526 PubChem SID number | 24859419 SMILES identifier | C(CCCCCCl)CCCCO InChI identifier | InChI=1/C10H21ClO/c11-9-7-5-3-1-2-4-6-8-10-12/h12H, 1-10H2 MDL number | MFCD00004748
CAS number | 51309-10-5 PubChem CID number | 123526 PubChem SID number | 24859419 SMILES identifier | C(CCCCCCl)CCCCO InChI identifier | InChI=1/C10H21ClO/c11-9-7-5-3-1-2-4-6-8-10-12/h12H, 1-10H2 MDL number | MFCD00004748

Safety properties

flash point | 110 °C
flash point | 110 °C